From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math Subject: Re: Eigenvalues and determinants. Date: 26 Nov 1998 20:56:00 GMT In article <365D8ECE.11863825@doc.ic.ac.uk>, ht97 writes: |> Today I was told that the determinant of a matrix is equal to the |> product of its eigenvalues. Could someone please supply a proof of |> this? Let P(t) = det(A - tI) be the characteristic polynomial of A. For an n x n matrix this is a polynomial in t of degree n, and leading term (-t)^n. Now (r is an eigenvalue of A) <=> (A - r I) is not invertible <=> P(r) = det(A - r I) = 0 <=> r - t is a factor of P(t). In particular, if A has n distinct eigenvalues, then P(t) has n distinct factors r - t, and since it has degree n and leading term (-t)^n, P(t) must be the product of the factors r - t over all the eigenvalues r. In particular, det(A) = P(0) is the product of the eigenvalues r. If the eigenvalues are not distinct, the product must be taken with multiplicities. The multiplicity m(r) of eigenvalue r is the highest power of r - t that divides P(t). We must have P(t) = product of (r-t)^m(r) over all eigenvalues r (by the Fundamental Theorem of Algebra every polynomial must factor into linear factors, and the only possible linear factors are of this form), and then again det(A) = P(0) = product of r^(m(r)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Newsgroups: sci.math Subject: Re: Eigenvalues and determinants. Date: 26 Nov 1998 16:15:03 -0500 In article <365D8ECE.11863825@doc.ic.ac.uk>, ht97 wrote: >Today I was told that the determinant of a matrix is equal to the >product of its eigenvalues. Could someone please supply a proof of >this? Hard to accommodate you without knowing more about your background knowledge. My path to proving this would be: find Schur decomposition of the matrix: A = U' * R * U, U unitary and R upper triangular, plus: U' is the inverse of U, plus: determinant of product is product of determinants. We know the eigenvalues of an upper triangular matrix, and hence the proof. Cheers. ZVK(Slavek).